Regarding the lab from Friday on how much heat the water absorbed when turning from ice to boiling water, there are three stages. The first stage is the melting of ice into water.the formula is q=ml. q is the heat, m is the mass and l is the latent heat value. The second stage is the water heating up which is the original q=mcΔt. c is the specific heat capacity and Δt is the change in temperature. There is a latent heat value for fusion (solid to liquid) and vaporization (liquid to gas). The negative values are used if it is the opposite (gas to liquid/liquid to solid).
To summarize, q=ml and latent heat values are used only for phase changes (change of state). Negative latent heat values are for phase changes towards solids.
THIS IS TESTABLE
Heat of a reaction is change in ENTHALPY. Change in enthalpy is the amount of heat produced or used during a chemical reaction. It is ΔH and has units of kJ/mol. ΔH is also equal to the different in enthalpies of products and reactants. Sometimes ΔH will have a subscript to indicate the type of reaction. ΔHc for combustion.
There are multiple ways to calculate ΔH 1. using the value of q and the number of moles 2. using Hess's law 3. using bond energies 4. Using standard heats of formation 5. Potential Energy Diagram
ΔH for exothermic reactions will be negative and endothermic reactions will be positive.
Using standard heats of formation. The value of ΔH can be calculated using the standard ethalpies of formation for the products and reactants. The standard enthalpy of formation (H°subscript f) is the energy associated with making a substance from its elements. Elements of a H° of 0 (O2 or H2) the values for compounds can be found on a table. A balanced chemical equation with states is need to calculateΔH. ΔH = ɛnH°f products - ɛnH°f reactants
Because bond breaking is an endothermic process and bond forming is an exothermic process, enthalpy of a reaction can be calculated using bond energies. ΔH = ɛ bonds broken - ɛ bonds formed Before adding, the bond energies must be multiplied by the number of that bond present AND the coefficient from the balanced chemical equation. When writing the balanced equation, the structure of each species must be drawn to determine the types of bonds present.
Hess's Law Hess's Law states that the heat of a reaction may be determined algebraically by adding the ΔH values for the related reactions. When added, these reference reactions must give the desired or target reaction. When determining ΔH, it may be necessary to reverse and/or adjust molar ratios by multiplying by a factor. Follow these rules when calculating for ΔH of the target reaction. 1. If a reaction must be reversed, the sing of the ΔH for that reaction must be changed (negative to positive and vice versa) 2. When adjusting molar ratios by multiplying, all coefficients in that equation must be multiplied by the same factor, as well as the ΔH value for that reaction. (basically, whatever is done to the equation must be done to its corresponding ΔH value.)
Ms. Wilson put up two videos that help explain Hess's Law as ISP work. Those videos can also help explain because I don't have visuals to put onto the wiki.
Representing Energy Changes Graphically Abinethaa Paramasivam| February 12, 2015 Summary of What we did Today- learned to show energy changes using a graph- learned about exothermic reactions and endothermic reactions The changes in energy for a chemical system can be represented graphically using a potential energy diagram. The reaction coordinate or reaction progress is the x-axis.These graphs plot potential energy vs. reaction progress. The potential energy is the y axis. Delta H is calculated by determining the difference in potential energy between the reactants and products. Endothermic Reactions
Exothermic Reactions
activation energy- minimum amount of energy required by reactants for reaction to occur Activated complex= collision state Ea in exothermic reaction is lower than an endothermic reaction. How do you calculate Ea for reverse? Answer: products to activated complex.
Example: Calculate heat produced by combustion of 2.4gram of ethane.
C2H6 (g) + 7/2 O2(g) à 2CO2 (g)+ 3H2O (g) Mass= 2.4g Molar Mass=30.08 g/mol N= 0.080mol Delta H= -1560kJ/mol 1mol/-1560kJ=0.080mol/x x= (0.080mol) (-1560kJ/mol) =-124.8kJ q=-124.8kJ Therefore the heat produced by combustion is -120kJ.
Homework Page 369 #17-18 Page 371 #12, 13, 15 Reminder: Quiz on Tuesday everything learned until Hess Law (this material is not on the Quiz)
Abinethaa Paramasivam|February 20, 2015 Summary of What we Learned today??? - learned about Maxwell-Boltzman diagrams - reviewed on drawing exothermic and endothermic reactions - Above graph is endothermic because product potential energy>reactants potential energy
Potential Energy Diagrams and Reaction Rate - lower activation energy greater, greater the rate of reaction - in general, exothermic reactions tend to have greater rates than endothermic reactions - many exothermic reactions are self-sustaining (can keep themselves going)
- if there is an increase in temperature, potential energy does not change - take this heat and have a vertical translation up; delta H does not change and activation energy does not change - if a catalyst is added, activation energy decreases - catalyst increases the rate by providing an alternate pathway mechanism with a lower activation energy
Maxwell-Boltzman Distributions To determine the total number of particles, that react then look at the patent area, calculate area in there and the number of particles that will react. - to increase rate, increase the area (by increasing the temperature)
Temperature changes shape of the graph because by increasing the temperature, you will increase the average kinetic energy..
Maxwell-Boltzman Distribution and Reaction Rates - increase rate of reaction, fraction of particles with activation energy or greater must increase. - increase red and purple (temperature is bigger) more particles with minimum energy - purple area=lower temp= lower area Homework: Read pages 366-367 Page 372 #11-16 Page 379 #9-10 Reminder: Unit test on Friday February 27 Formal Lab report due for heat lab due on Thursday February 26
There is a mathematical relationship between the rate of the reaction and the factors affecting the reaction
This relationship must be determined empirically (must analyse experimental data)
The rate law for a reaction describes the relationship between rate (r) and the product of the initial concentrations of the reactants raised to some exponential values
GIVEN aX + bY --> products r α [X]^m [Y]^n
The exponents m and n describe the relationship between rate and initial concentration and must be determined experimentally
The value of m and n may be a whole number, zero, or a fraction and do not have to equal the coefficients from the balanced chemical equation (ie a and b)
To determine the rate of a reaction mathematically, a constant k called the rate constant must be introduced into the relationship
The rate law equation for the reaction is written as r = k [X]^m [Y]^n
The value for the rate constant must also be determined experimentally
ORDERS OF REACTIONS
The exponents of the rate law equation are called the orders of reaction
The sum of all the individual orders (exponents) is referred to as the overall order of reaction
GIVEN 2X + 2Y + 3Z --> products r = k [X]^1 [Y]^2 [Z]^0
The order of reaction for X is 1, Y is 2, and Z is 0 SO the overall order is 3 (1+2+0=3)
WHAT DOES THE ORDER OF REACTION MEAN
For 1st order, if the concentration is doubled, the rate is doubled {2^1}
For 2st order, if the concentration is doubled, the rate is quadrupled {2^2}
For 3rd order, if the concentration is doubled, the rate is increased by a factor of 8 {2^3}
For zero order, if the concentration is doubled, there is no change to the rate {2^0}
If a reaction is zero order for one of the reactants, that reactant is left out of the rate equation
SAMPLE RATE PROBLEM
The data below was collected for the reaction
A(aw) + 2B(aq) + 3C(aq) --> E(s) + F(aq)
Trial #
[A] (mol/L)
[B] (mol/L)
[C] (mol/L)
Initial rate (mol/Ls)
1
0.10
0.10
0.10
4.0
2
0.10
0.20
0.10
16.0
3
0.40
0.20
0.10
32.0
4
0.10
0.10
0.30
12.0
Determine the rate law equation, including the k value:
For A use 2 and 3
[A] x 4
r x 2 = 4^1/2 ∴ ½ order
For B use 1 and 2
[B] x 2
r x 4 = 2^2 ∴ 2nd order
For C use 1 and 4
[C] x 3
r x 3 = 3 ∴ 1st order
r = k [A]^1/2 [B]^2 [C] Use trial 4 to solve for k (but you can use any trial) 12.0 = k {0.10}^1/2 {0.10}^2 {0.30} k = 12649 = 1.3 x 10^4 ?units?
?How to find units? r = k [A]^1/2 [B]^2 [C]
mol/Ls = k (mol^1/2/L^1/2) (mol^2/L^2) (mol/L)
mol/Ls = k (mol^7/2/L^7/2)
k = (mol/Ls) (L^7/2/mol^7/2)
k = L^2.5/mol^2.5s
Easier Method to Find Units
k (units) = L^(n-1) / mol^(n-l) * s
n represents the value for the overall order
Lower molecular steps tend to be faster than ones with more molecules
That rate determining step is the slowest step in the reaction mechanism and will have the largest activation energy
USING REACTION MECHANISMS TO DETERMINE THE RATE LAW EQUATION
The rate law expression can be determined from the reaction mechanism
The rate determining step is the elementary step that determines the rate law expression
The rate of a reaction is proportional to the concentration of the reactants in the rate determining step raised to their molar coefficients
If any reaction intermediates appear in this expression, you must use the other steps to eliminate these species
Only reversible equations, denoted by two arrows pointing in opposite directions, can be used to replace reaction intermediates
Example 1:
The reaction A + 2B --> C + D occurs by the mechanism below. Use that mechanism to determine the rate law expression.
Step 1: A + B <--> AB (fast) Note: <--> represents reversibility
Step 2: AB + B --> C + D (slow)
Equation: A + 2B --> C+ D (cancel out chemicals that are on both sides of step equations to verify the overall equation)
Use step 2 because its the RDS (rate determining step/slope) (its the slowest)
r = k [AB] [B] AB is a reaction intermediate so use step 1 to replace AB
r1 = [A] [B] = [AB] (because its a reversible step therefore rate of the forward = rate of the reverse. Also, k is ignored in this step) Isolate [AB] = [A] [B] and sub into above equation for [AB]
r = k [AB] [B]
r = k [A] [B] [B]
r = k [A] [B]^2 (final rate law expression)
Example 2:
The reaction between NO and H2 is believed to occur in the following three-step process. Write the overall equation for this reaction and the rate law expression.
Step 1: 2NO <--> N2O2 {fast)
Step 2: N2O2 + H2 --> N2O + H2O (slow)
Step 3: N2O + H2 ---> N2 + H2O (fast)
Overall equation: 2NO + 2H2 ---> N2 + 2H2O
Use step 2 because its the RDS
r = k [N2O2] [H2] Use step 1 (it has N2O2 and is reversible)
r1 = [NO]^2 = [N2O2]
r = k [NO]^2 [H2] (final rate law expression)
Phase Change and Heat
Regarding the lab from Friday on how much heat the water absorbed when turning from ice to boiling water, there are three stages. The first stage is the melting of ice into water.the formula is q=ml. q is the heat, m is the mass and l is the latent heat value. The second stage is the water heating up which is the original q=mcΔt. c is the specific heat capacity and Δt is the change in temperature. There is a latent heat value for fusion (solid to liquid) and vaporization (liquid to gas). The negative values are used if it is the opposite (gas to liquid/liquid to solid).To summarize, q=ml and latent heat values are used only for phase changes (change of state). Negative latent heat values are for phase changes towards solids.
THIS IS TESTABLE
Heat of a reaction is change in ENTHALPY. Change in enthalpy is the amount of heat produced or used during a chemical reaction. It is ΔH and has units of kJ/mol. ΔH is also equal to the different in enthalpies of products and reactants. Sometimes ΔH will have a subscript to indicate the type of reaction. ΔHc for combustion.
There are multiple ways to calculate ΔH
1. using the value of q and the number of moles
2. using Hess's law
3. using bond energies
4. Using standard heats of formation
5. Potential Energy Diagram
ΔH for exothermic reactions will be negative and endothermic reactions will be positive.
Using standard heats of formation. The value of ΔH can be calculated using the standard ethalpies of formation for the products and reactants. The standard enthalpy of formation (H° subscript f) is the energy associated with making a substance from its elements. Elements of a H° of 0 (O2 or H2) the values for compounds can be found on a table. A balanced chemical equation with states is need to calculateΔH.
ΔH = ɛnH°f products - ɛnH°f reactants
H°f must be multiplied by its molar coefficient.
https://www.youtube.com/watch?v=6KnoNK8fahM
http://www.kentchemistry.com/links/Matter/PhaseChanges.htm
Calculating Enthalpy Change
Because bond breaking is an endothermic process and bond forming is an exothermic process, enthalpy of a reaction can be calculated using bond energies.ΔH = ɛ bonds broken - ɛ bonds formed
Before adding, the bond energies must be multiplied by the number of that bond present AND the coefficient from the balanced chemical equation. When writing the balanced equation, the structure of each species must be drawn to determine the types of bonds present.
This is a link to a video that will also explain how to use bond energies.
https://www.youtube.com/watch?v=ih11AH0lEZY
Hess's Law
Hess's Law states that the heat of a reaction may be determined algebraically by adding the ΔH values for the related reactions. When added, these reference reactions must give the desired or target reaction. When determining ΔH, it may be necessary to reverse and/or adjust molar ratios by multiplying by a factor.
Follow these rules when calculating for ΔH of the target reaction.
1. If a reaction must be reversed, the sing of the ΔH for that reaction must be changed (negative to positive and vice versa)
2. When adjusting molar ratios by multiplying, all coefficients in that equation must be multiplied by the same factor, as well as the ΔH value for that reaction.
(basically, whatever is done to the equation must be done to its corresponding ΔH value.)
Ms. Wilson put up two videos that help explain Hess's Law as ISP work. Those videos can also help explain because I don't have visuals to put onto the wiki.
https://www.youtube.com/watch?v=wH0jn0qb3J4
https://www.youtube.com/watch?v=9oaNmEqY-ss
Representing Energy Changes Graphically
Abinethaa Paramasivam| February 12, 2015
Summary of What we did Today- learned to show energy changes using a graph- learned about exothermic reactions and endothermic reactions
The changes in energy for a chemical system can be represented graphically using a potential energy diagram. The reaction coordinate or reaction progress is the x-axis. These graphs plot potential energy vs. reaction progress. The potential energy is the y axis. Delta H is calculated by determining the difference in potential energy between the reactants and products.
Endothermic Reactions
Exothermic Reactions
activation energy- minimum amount of energy required by reactants for reaction to occur
Activated complex= collision state
Ea in exothermic reaction is lower than an endothermic reaction. How do you calculate Ea for reverse? Answer: products to activated complex.
Example: Calculate heat produced by combustion of 2.4gram of ethane.
C2H6 (g) + 7/2 O2(g) à 2CO2 (g)+ 3H2O (g)
Mass= 2.4g
Molar Mass=30.08 g/mol
N= 0.080mol
Delta H= -1560kJ/mol
1mol/-1560kJ=0.080mol/x
x= (0.080mol) (-1560kJ/mol)
=-124.8kJ
q=-124.8kJ
Therefore the heat produced by combustion is -120kJ.
Homework
Page 369 #17-18
Page 371 #12, 13, 15
Reminder: Quiz on Tuesday everything learned until Hess Law (this material is not on the Quiz)
Watch these videos to review for Tuesday's quiz:
https://www.youtube.com/watch?v=f_jCFXoRX4w&index=10&list=WL
https://www.youtube.com/watch?v=JuWtBR-rDQk
Udari Premachandra | February 17, 2015
Rates of Reaction:
Calculating Average Rate of Reaction:
Homework:
REMEMBER: ISP FEBRUARY 18!
Links to Other Resources:
Chemistry 11.3 Rate of Reaction and Rate LawsChemical Reaction Rate - Intro
Udari Premachandra | February 19, 2015
Heat of Reaction Lab was completed today.
[[#Udari Premachandra | February 19, 2015-Homework:]]Homework:
[[#Udari Premachandra | February 19, 2015-Links to other Resources:]]Links to other Resources:
Moles Calculations - Reacting MassesHydrochloric acid : sodium hydroxide
Potential Energy Diagrams
Abinethaa Paramasivam|February 20, 2015
Summary of What we Learned today???
- learned about Maxwell-Boltzman diagrams
- reviewed on drawing exothermic and endothermic reactions
- Above graph is endothermic because product potential energy>reactants potential energy
Potential Energy Diagrams and Reaction Rate
- lower activation energy greater, greater the rate of reaction
- in general, exothermic reactions tend to have greater rates than endothermic reactions
- many exothermic reactions are self-sustaining (can keep themselves going)
- if there is an increase in temperature, potential energy does not change
- take this heat and have a vertical translation up; delta H does not change and activation energy does not change
- if a catalyst is added, activation energy decreases
- catalyst increases the rate by providing an alternate pathway mechanism with a lower activation energy
Maxwell-Boltzman Distributions
To determine the total number of particles, that react then look at the patent area, calculate area in there and the number of particles that will react.
- to increase rate, increase the area (by increasing the temperature)
Temperature changes shape of the graph because by increasing the temperature, you will increase the average kinetic energy..
Maxwell-Boltzman Distribution and Reaction Rates
- increase rate of reaction, fraction of particles with activation energy or greater must increase.
- increase red and purple (temperature is bigger)
more particles with minimum energy
- purple area=lower temp= lower area
Homework: Read pages 366-367
Page 372 #11-16
Page 379 #9-10
Reminder: Unit test on Friday February 27
Formal Lab report due for heat lab due on Thursday February 26
If you need more help, watch these videos:
https://www.youtube.com/watch?v=mGt73Eg0hjE
https://www.youtube.com/watch?v=48sLH9P8QK0
==
==
Rate Laws and Order of Reaction
February 23, 2015Xinli Tu
GIVEN aX + bY --> products
r α [X]^m [Y]^n
ORDERS OF REACTIONS
GIVEN 2X + 2Y + 3Z --> products
r = k [X]^1 [Y]^2 [Z]^0
WHAT DOES THE ORDER OF REACTION MEAN
SAMPLE RATE PROBLEM
The data below was collected for the reaction
A(aw) + 2B(aq) + 3C(aq) --> E(s) + F(aq)
For A use 2 and 3
[A] x 4
r x 2 = 4^1/2
∴ ½ order
For B use 1 and 2
[B] x 2
r x 4 = 2^2
∴ 2nd order
For C use 1 and 4
[C] x 3
r x 3 = 3
∴ 1st order
r = k [A]^1/2 [B]^2 [C]
Use trial 4 to solve for k (but you can use any trial)
12.0 = k {0.10}^1/2 {0.10}^2 {0.30}
k = 12649 = 1.3 x 10^4 ?units?
?How to find units?
r = k [A]^1/2 [B]^2 [C]
mol/Ls = k (mol^1/2/L^1/2) (mol^2/L^2) (mol/L)
mol/Ls = k (mol^7/2/L^7/2)
k = (mol/Ls) (L^7/2/mol^7/2)
k = L^2.5/mol^2.5s
Easier Method to Find Units
k (units) = L^(n-1) / mol^(n-l) * s
n represents the value for the overall order
∴ r = 1.3 x 10^4 L^2.5/mol^2.5s [A]^1/2 [B]^2 [C]
Resource
https://www.youtube.com/watch?v=tSI8vcM-9Kw
http://www.sparknotes.com/chemistry/kinetics/ratelaws/section2.rhtml
Homework
Reminders
Reaction Mechanisms
February 24, 2015Xinli Tu
USING REACTION MECHANISMS TO DETERMINE THE RATE LAW EQUATION
Example 1:
The reaction A + 2B --> C + D occurs by the mechanism below. Use that mechanism to determine the rate law expression.
Step 1: A + B <--> AB (fast) Note: <--> represents reversibility
Step 2: AB + B --> C + D (slow)
Equation: A + 2B --> C+ D (cancel out chemicals that are on both sides of step equations to verify the overall equation)
Use step 2 because its the RDS (rate determining step/slope) (its the slowest)
r = k [AB] [B]
AB is a reaction intermediate so use step 1 to replace AB
r1 = [A] [B] = [AB] (because its a reversible step therefore rate of the forward = rate of the reverse. Also, k is ignored in this step)
Isolate [AB] = [A] [B] and sub into above equation for [AB]
r = k [AB] [B]
r = k [A] [B] [B]
r = k [A] [B]^2 (final rate law expression)
Example 2:
The reaction between NO and H2 is believed to occur in the following three-step process. Write the overall equation for this reaction and the rate law expression.
Step 1: 2NO <--> N2O2 {fast)
Step 2: N2O2 + H2 --> N2O + H2O (slow)
Step 3: N2O + H2 ---> N2 + H2O (fast)
Overall equation: 2NO + 2H2 ---> N2 + 2H2O
Use step 2 because its the RDS
r = k [N2O2] [H2]
Use step 1 (it has N2O2 and is reversible)
r1 = [NO]^2 = [N2O2]
r = k [NO]^2 [H2] (final rate law expression)
Resource
https://www.youtube.com/watch?v=6OKZ2S6FnYc
http://www.brightstorm.com/science/chemistry/chemical-reaction-rates/reaction-mechanism/
Homework
Reminders
Alka Seltzer Lab
February 26th, 2015Vishal Sharma
Today we did the Alka Seltzer lab. You must do a ONE page lab report.
Lab Report can be found at: https://sch4uking.wikispaces.com/Lab+Reports
Alka Seltzer Lab: https://www.youtube.com/watch?v=TKpMTz06qrU
Factors Affecting Rates of Reaction: https://www.youtube.com/watch?v=S1sWvCOTUl8
Homework:
This lab report is due on March 5th, 2015
Study for Unit test tomorrow